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## Decisions, decisions

Newcomb’s paradox is the name usually given to the following problem. You are playing a game against another player, often called Omega, who claims to be omniscient; in particular, Omega claims to be able to predict how you will play in the game. Assume that Omega has convinced you in some way that it is, if not omniscient, at least remarkably accurate: for example, perhaps it has accurately predicted your behavior many times in the past.

Omega places before you two opaque boxes. Box A, it informs you, contains $1,000. Box B, it informs you, contains either$1,000,000 or nothing. You must decide whether to take only Box B or to take both Box A and Box B, with the following caveat: Omega filled Box B with \$1,000,000 if and only if it predicted that you would take only Box B.

What do you do?

(If you haven’t heard this problem before, please take a minute to decide on an option before continuing.)

## The double commutant theorem

Let $A$ be an abelian group and $T = \{ T_i : A \to A \}$ be a collection of endomorphisms of $A$. The commutant $T'$ of $T$ is the set of all endomorphisms of $A$ commuting with every element of $T$; symbolically,

$\displaystyle T' = \{ S \in \text{End}(A) : TS = ST \}$.

The commutant of $T$ is equal to the commutant of the subring of $\text{End}(A)$ generated by the $T_i$, so we may assume without loss of generality that $T$ is already such a subring. In that case, $T'$ is just the ring of endomorphisms of $A$ as a left $T$-module. The use of the term commutant instead can be thought of as emphasizing the role of $A$ and de-emphasizing the role of $T$.

The assignment $T \mapsto T'$ is a contravariant Galois connection on the lattice of subsets of $\text{End}(A)$, so the double commutant $T \mapsto T''$ may be thought of as a closure operator. Today we will prove a basic but important theorem about this operator.

## Walks on graphs and tensor products

Recently I asked a question on MO about some computations I’d done with Catalan numbers awhile ago on this blog, and Scott Morrison gave a beautiful answer explaining them in terms of quantum groups. Now, I don’t exactly know how quantum groups work, but along the way he described a useful connection between walks on graphs and tensor products of representations which at least partially explains one of the results I’d been wondering about and also unites several other related computations that have been on my mind recently.

Let $G$ be a compact group and let $\text{Rep}(G)$ denote the category of finite-dimensional unitary representations of $G$. As in the finite case, due to the existence of Haar measure, $\text{Rep}(G)$ is semisimple (i.e. every unitary representation decomposes uniquely into a sum of irreducible representations), and via the diagonal action it comes equipped with a tensor product with the property that the character of the tensor product is the product of the characters of the factors.

Question: Fix a representation $V \in \text{Rep}(G)$. What is the multiplicity of the trivial representation in $V^{\otimes n}$?